How will you find the page with most incoming links from billions of
web-pages
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i found this website from my friend
On Thu, Aug 12, 2010 at 11:41 AM, vijay auvija...@gmail.com wrote:
How will you find the page with most incoming links from billions of
web-pages
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To
http://www.hpl.hp.com/personal/Vinay_Deolalikar/Papers/pnp12pt.pdf
Rahul K Rai
rahulpossi...@gmail.com
2010/8/12 Avik Mitra tutai...@gmail.com
We are proud that an Indian has attempted to solve this problem.
Avik
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Construct a graph of these pages
Run a BFS/DFS.
The vertex with maximum incoming connections would be the page which is
referenced most.
On Thu, Aug 12, 2010 at 11:41 AM, vijay auvija...@gmail.com wrote:
How will you find the page with most incoming links from billions of
web-pages
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You
:-) that was funny.
I think you can take a look at the concepts of web-page ranking algorithms
- http://www.cpccci.com/academics/surveypagerank.htm
http://www.cpccci.com/academics/surveypagerank.htmKishen
On Thu, Aug 12, 2010 at 1:33 AM, thiru pujari thiru.puj...@gmail.comwrote:
i found this
How to find two missing numbers from an unsorted continuous natural
numbers array [only use O(1) space and O(n) time]
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To
take the sum of array and product.
if they were present then sum will be n(n-1)/2 and product will be n!.
make two eq of a+b and a-b with these values and get the num
On Thu, Aug 12, 2010 at 5:26 AM, vijay auvija...@gmail.com wrote:
How to find two missing numbers from an unsorted continuous
@ashish. The product will overflow for even moderate n, so instead,
form the sum and the sum of the squares of the numbers. If a and b are
the missing numbers, they satisfy
a + b = n(n+1)/2 - sum of the numbers
a^2 + b^2 = n(n+1)(2n+1)/6 - sum of the squares of the numbers.
Solve by the method
Johnny was asked by his math teacher to compute nn (n to the power of
n, where n is an integer), and has to read his answer out loud. This
is a bit of a tiring task, since the result is probably an extremely
large number, and would certainly keep Johnny occupied for a while if
he were to do it
